Download Lecture 2.2 - Sybil Nelson

2
Probability
Copyright © Cengage Learning. All rights reserved.
2.2
Axioms, Interpretations,
and Properties of Probability
Copyright © Cengage Learning. All rights reserved.
Axioms, Interpretations, and Properties of Probability
You might wonder why the third axiom contains no
reference to a finite collection of disjoint events.
3
Axioms, Interpretations, and Properties of Probability
Proposition
4
Example 2.11
Consider tossing a thumbtack in the air. When it comes to
rest on the ground, either its point will be up (the outcome U)
or down (the outcome D). The sample space for this event is
therefore = {U, D}.
The axioms specify P( ) = 1, so the probability assignment
will be completed by determining P(U) and P(D).
Since U and D are disjoint and their union is , the foregoing
proposition implies that
1 = P(
) = P(U) + P(D)
5
Interpreting Probability
6
Interpreting Probability
Relative Frequency
Consider flipping a coin. We all know that the probability of
getting heads and the probability of getting tails is 50%.
Student A performs 10 coin flips and gets 7 heads
Student B performs 10 coin flips and gets 2 heads
Why didn’t they both get 5 heads?
7
More Probability Properties
8
More Probability Properties
Proposition
9
More Probability Properties
In general, the foregoing proposition is useful when the
event of interest can be expressed as “at least . . . ,” since
then the complement “less than . . .” may be easier to work
with (in some problems, “more than . . .” is easier to deal
with than “at most . . .”).
When you are having difficulty calculating P(A) directly,
think of determining P(A).
10
Problem 15
Consider the type of clothes dryer (gas or electric)
purchased by each of five different customers at a certain
store.
a. If the probability that at most one of these purchases an
electric dryer is .428, what is the probability that at least
two purchase an electric dryer?
b. If P(all five purchase gas)=.116 and P(all five purchase
electric)=.005, what is the probability that at least one of
each type is purchased?
11
More Probability Properties
Proposition
This is because 1 = P(A) + P(A)  P(A) since P(A)  0.
When events A and B are mutually exclusive,
P(A  B) = P(A) + P(B).
Proposition
12
More Probability Properties
Proof
Note first that A ∪ B can be decomposed into two disjoint
events, A and B ∩ A’; the latter is the part of B that lies
outside A (see Figure 2.4). Furthermore, B itself is the
union of the two disjoint events A ∩ B and A’ ∩ B, so P(B) =
P(A ∩ B) 1 P(A’ ∩ B). Thus
13
More Probability Properties
The addition rule for a triple union probability is similar to
the foregoing rule.
14
More Probability Properties
This can be verified by examining a Venn diagram of
A  B  C, which is shown in Figure 2.6.
ABC
Figure 2.6
When P(A), P(B), and P(C) are added, the intersection
probabilities P(A ∩ B), P(A ∩ C), and P(B ∩ C) are all
counted twice. Each one must therefore be subtracted.
But then P(A ∩ B ∩ C) has been added in three times and
subtracted out three times, so it must be added back.
15
Determining Probabilities
Systematically
16
Determining Probabilities Systematically
Consider a sample space that is either finite or “countably
infinite” (the latter means that outcomes can be listed in an
infinite sequence, so there is a first outcome, a second
outcome, a third outcome, and so on—for example, the
battery testing scenario of Example 12).
Let E1, E2, E3, … denote the corresponding simple events,
each consisting of a single outcome.
17
Determining Probabilities Systematically
A sensible strategy for probability computation is to first
determine each simple event probability, with the
requirement that P(Ei) = 1.
Then the probability of any compound event A is computed
by adding together the P(Ei)’s for all Ei’s in A:
18
Example 2.15
During off-peak hours a commuter train has five cars.
Suppose a commuter is twice as likely to select the middle
car (#3) as to select either adjacent car (#2 or #4), and is
twice as likely to select either adjacent car as to select
either end car (#1 or #5).
Let pi = P(car i is selected) = P(Ei). Then we have
p3 = 2p2 = 2p4 and p2 = 2p1 = 2p5 = p4. This gives
1 = P(Ei) = p1 + 2p1 + 4p1 + 2p1 + p1 = 10p1
implying p1 = p5 = .1, p2 = p4 = .2, p3 = .4. The probability
that one of the three middle cars is selected (a compound
event) is then p2 + p3 + p4 = .8.
19
Equally Likely Outcomes
20
Equally Likely Outcomes
In many experiments consisting of N outcomes, it is
reasonable to assign equal probabilities to all N simple
events.
These include such obvious examples as tossing a fair coin
or fair die once or twice (or any fixed number of times), or
selecting one or several cards from a well-shuffled deck
of 52. With p = P(Ei) for every i,
That is, if there are N equally likely outcomes, the
probability for each is
21
Equally Likely Outcomes
Now consider an event A, with N(A) denoting the number of
outcomes contained in A. Then
Thus when outcomes are equally likely, computing
probabilities reduces to counting: determine both the
number of outcomes N(A) in A and the number of
outcomes N in , and form their ratio.
22
Example 2.16
You have six unread mysteries on your bookshelf and six
unread science fiction books.
The first three of each type are hardcover, and the last
three are paperback.
Consider randomly selecting one of the six mysteries and
then randomly selecting one of the six science fiction books
to take on a post-finals vacation to Acapulco (after all, you
need something to read on the beach).
Number the mysteries 1, 2, . . . , 6, and do the same for the
science fiction books.
23
Example 2.16
cont’d
Then each outcome is a pair of numbers such as (4, 1),
and there are N = 36 possible outcomes (For a visual of
this situation, refer the table below and delete the first row
and column).
24
Example 2.16
cont’d
With random selection as described, the 36 outcomes are
equally likely.
Nine of these outcomes are such that both selected books
are paperbacks (those in the lower right-hand corner of the
referenced table): (4, 4), (4, 5), . . . , (6, 6).
So the probability of the event A that both selected books
are paperbacks is
25